Ch. 5 Notes---Scientific Measurement
Qualitative vs. Quantitative
•
Qualitative measurements give results in a descriptive nonnumeric
form. (The result of a measurement is an _____________
adjective
describing the object.)
short
heavy long, __________...
cold
*Examples: ___________,
___________,
•
Quantitative measurements give results in numeric form. (The
number
results of a measurement contain a _____________.)
600 lbs. 22 meters, __________...
5 ºC
*Examples: 4’6”, __________,
Accuracy vs. Precision
•
single measurement is to the
Accuracy is how close a ___________
true __________
value
________
of whatever is being measured.
•
several measurements are to
Precision is how close ___________
each ___________.
other
_________
Practice Problem: Describe the shots for the targets.
Bad Accuracy & Bad Precision
Good Accuracy & Bad Precision
Bad Accuracy & Good Precision
Good Accuracy & Good Precision
Significant Figures
•
Significant figures are used to determine the ______________
of a
precision
measurement. (It is a way of indicating how __________
precise a
measurement is.)
*Example: A scale may read a person’s weight as 135 lbs. Another
scale may read the person’s weight as 135.13 lbs. The ___________
second
more significant figures in the
scale is more precise. It also has ______
measurement.
•
•
•
Whenever you are measuring a value, (such as the length of an
object with a ruler), it must be recorded with the correct number of
sig. figs.
ALL the numbers of the measurement known for sure.
Record ______
Record one last digit for the measurement that is estimated. (This
reading in between the
means that you will be ________________________________
marks of the device and taking a __________
guess
__________
at what the next
number is.)
Significant Figures
•
Practice Problems: What is the length recorded to the correct
number of significant figures?
length = ________cm
11.65
(cm) 10
20
30
40
length = ________cm
58
50
60
70
80
90
100
For Example
•
•
•
Lets say you are finding the average mass of beans. You would
count how many beans you had and then find the mass of the
beans.
26 beans have a mass of 44.56 grams.
44.56 grams ÷26 =1.713846154 grams
So then what should your written answer be?
How many decimal points did you have in
your measurement?
2
Rounded answer = 1.71 grams
•
•
The SI System (The Metric System)
Here is a list of common units of measure used in science:
Standard Metric Unit
Quantity Measured
mass
kilogram, (gram)
______________
length
meter
______________
cubic meter, (liter)
______________
volume
seconds
______________
time
temperature
Kelvin, (˚Celsius)
_____________
The following are common approximations used to convert from our
English system of units to the metric system:
1 yard
1 m ≈ _________
2.2 lbs.
1 kg ≈ _______
1.609 km ≈ 1 mile
mass of a small paper clip
1 gram ≈ ______________________
sugar cube’s volume
1mL ≈ _____________
1 L ≈ 1.06 quarts
dime
1mm ≈ thickness of a _______
The SI System (The Metric System)
•
Metric Conversions
The metric system prefixes are based on factors of _______.
mass Here is a
list of the common prefixes used in chemistry:
kilo- hecto- deka-
•
•
deci- centi- milli-
The box in the middle represents the standard unit of measure such
as grams, liters, or meters.
Moving from one prefix to another involves a factor of 10.
cm = 10 _____
dm = 1 _____
m
*Example: 1000 millimeters = 100 ____
•
The prefixes are abbreviated as follows:
k h da g, L, m d c m
grams
Liters
meters
*Examples of measurements: 5 km 2 dL 27 dag 3 m 45 mm
Metric Conversions
•
To convert from one prefix to another, simply count how many places
you move on the scale above, and that is the same # of places the
decimal point will move in the same direction.
deci- centi- milliPractice Problems: kilo- hecto- deka380,000
0.00145
380 km = ______________m
1.45 mm = _________m
461 mL = ____________dL
4.61
0.4 cg = ____________
0.0004
dag
0.26 g =_____________
mg
230,000 m = _______km
260
230
Other Metric Equivalents
1 mL = 1 cm3
1 L = 1 dm3
For water only:
1 L = 1 dm3 = 1 kg of water or 1 mL = 1 cm3 = 1 g of water
Practice Problems:
0.3 L
(1) How many liters of water are there in 300 cm3 ? ___________
50 kg
(2) How many kg of water are there in 500 dL? _____________
Metric Volume: Cubic Meter (m3)
10 cm x 10 cm x 10 cm = Liter
Ch. 4 Problem Solving in Chemistry
Dimensional Analysis
conversion
• Used in _______________
problems.
*Example: How many seconds are there in 3 weeks?
• A method of keeping track of the_____________.
units
Conversion Factor
ratio of units that are _________________
equivalent
• A ________
to one another.
*Examples:
1 min/ ___
60 sec (or ___
60 sec/ 1 min)
7 days/ 1 week (or 1 week/ ___
7 days)
___
1000 m/ ___
1 km
(or ___
1 km/ 1000 m)
• Conversion factors need to be set up so that when multiplied, the unit
of the “Given” cancel out and you are left with the “Unknown” unit.
top and the
• In other words, the “Unknown” unit will go on _____
“Given” unit will go on the ___________
bottom
of the ratio.
How to Use Dimensional Analysis to Solve Conversion Problems
• Step 1:
Identify the “________”.
Given
This is typically the only number
given in the problem. This is your starting point. Write it down! Then
write “x _________”. This will be the first conversion factor ratio.
• Step 2:
Identify the “____________”.
This is what are you trying to
Unknown
figure out.
• Step 3:
Identify the ____________
Sometimes you will
conversion _________.
factors
simply be given them in the problem ahead of time.
• Step 4:
By using these conversion factors, begin planning a solution
to convert from the given to the unknown.
• Step 5:
When your conversion factors are set up, __________
multiply all the
divide
numbers on top of your ratios, and ____________
by all the numbers
on bottom.
If your units did not ________
cancel ______
out correctly, you’ve messed up!
Practice Problems:
(1)How many hours are there in 3.25 days?
3.25 days x 24 hrs = 78 hrs
1 day
(2) How many yards are there in 504 inches?
504 in. x 1 ft
12 in.
x 1 yard
3 ft
= 14 yards
(3) How many days are there in 26,748 seconds?
26,748 sec x 1 min x 1 hr x 1 day
60 sec 60 min
24 hrs
= 0.30958 days
Converting Complex Units
• A complex unit is a measurement with a unit in the _____________
numerator
and ______________.
denominator
*Example: 55 miles/hour 17 meters/sec 18 g/mL
• To convert complex units, simply follow the same procedure as
top first. Then convert the
before by converting the units on ______
bottom
units on __________
next.
Practice Problems: (1) The speed of sound is about 330 meters/sec.
What is the speed of sound in units of miles/hour? (1609 m = 1 mile)
330m x 1 mile x 3600 sec = 738 miles/hr
sec
1609 m
1 hr
(2) The density of water is 1.0 g/mL. What is the density of water in
units of lbs/gallon? (2.2 lbs = 1 kg) (3.78 L = 1 gal)
1.0 g x 1 kg x 2.2 lbs x 1000 mL x 3.78 L = 8.3 lbs/gal
mL
1000 g
1 kg
1L
1 gal
Ch. 6 Notes -- Chemical Composition
What is a mole?
Ch 6 – Chemical Quantities
The Mole!!!
• A counting unit
• Similar to a dozen, except instead of 12, it’s
602 billion trillion…
(602,000,000,000,000,000,000,000)
23
6.02
x
10
• ___________ (in scientific notation)
• This number is named in honor of Amedeo
_________
Avogadro (1776 – 1856), who studied quantities
of gases and discovered that no matter what the
gas was, there were the same number of
molecules present…6.02 x 1023
Just How Big is a Mole?
• Enough soft drink cans to cover the
surface of the earth to a depth of over 200
miles.
• If you had Avogadro's number of unpopped popcorn kernels, and spread them
across the United States of America, the
country would be covered in popcorn to a
depth of over 9 miles.
• If we were able to count atoms at the rate
of 10 million per second, it would take
about 2 billion years to count the atoms in
one mole.
The Mole
12 cookies
• 1 dozen cookies = ___
6.02 X 1023
• 1 mole of cookies = ___________
cookies
12 cars
• 1 dozen cars = ___
• 1 mole of cars = __________
6.02 X 1023 cars
12 Al atoms
• 1 dozen Al atoms = ___
6.02 X 1023 atoms
• 1 mole of Al atoms = __________
Note that the NUMBER is always the same, but
MASS is very different!
the ______
mol
Mole is abbreviated ______
.
The Mole and Mass
the sum
• Mass in grams of 1 mole equal to __________
of the atomic masses.
Practice problem:
Calculate the mass of 1 mole of CaCl2.
40.1 g/mol
Ca = 1 x ________
= 40.1 g/mol
35.5 g/mol
Cl = 2 x ________
= 71.0 g/mol
40.1 g/mol + 71.0 g/mol = __________
111.1
g/mol CaCl2
1 mole of CaCl2 = 111.1 g/mol
Mole Conversion Factors
that you will need to know!
6.02 x 1023 atoms/molecules/etc.
• 1 mol = __________
? ( molar mass) grams
• 1 mol = _____________
22.4
• 1 mol = ________
Liters of gas at STP
(STP is Standard Temp. and Pressure…
we will talk about what this means later!)
Ch. 6 Notes -- Chemical Composition
Practice Problems:
(1) How many atoms of hydrogen are there in each compound?
a) Ca(OH)2 ___
2
b) C3H8O___
8 c) (NH4)2HPO4 ___
9 d) HC2H3O2 ___
4
(2) Calculate the formula mass of each compound. (Add up all the
atomic masses for each atom from the Periodic Table.)
a) CaCO3
b) (NH4)2SO4
Ca = 40.1
C = 12.0
3 O’s =3 x 16.0 = 48.0
Add them up!
100.1 g/mol
c) C3H6O
C = 3 x 12.0 = 36.0
Add them up!
H = 6 x 1.0 = 6.0
58.0 g/mol
O =16.0
e) H3PO4 3 H’s = 3 x 1.0 = 3.0
P = 31.0
4 O’s = 4 x 16.0 =64.0
2 N’s = 2 x 14.0 = 28.0
8 H’s = 8 x 1.0 = 8.0
S = 32.1
4 O’s = 4 x 16.0 = 64.0
Add them up!
132.1 g/mol
d) Br2
2 Br’s = 2 x 79.9 = 159.8 g/mol
f) N2O5
Add them up!
98.0 g/mol
2 N’s = 2 x 14.0 = 28.0
5 O’s = 5 x 16.0 = 80.0
Add them up!
108.0 g/mol
3) Convert 835 grams of SO3 to moles.
835 g SO3 x
1 mole SO3 = 10.4 moles of SO3
80.1 g SO3
4) How many molecules of CH4 are there in 18 moles?
23 molecules CH
6.02
x
10
4 = 1.08 x 1025 molecules CH
18 moles CH4 x
4
1 mole CH4
5) How many grams of helium are there in 5.6 x 1023 atoms of helium?
5.6 x 1023 atoms He x
4.0 grams He
=
23
6.02 x 10 atoms He
3.72 grams He
6) How many molecules are there in 3.7 grams of H2O?
23
3.7 grams H2O x 6.02 x 10 molecules H2O = 1.24 x 1023 molecules H2O
18.0 grams H2O
Calculating Percent Composition by Mass
Step 1: Find the formula mass of the compound by adding the
individual masses of the elements together.
Step 2: Divide each of the individual masses of the elements by the
formula mass of the compound.
Step 3: Convert the decimal to a % by multiplying by 100.
Practice Problems:
(1) Find the % composition of the elements in each compound.
a) Na3PO4
3 Na’s = 3 x 23.0 = 69.0 ÷ 164 = 0.421 = 42.1%
P = 31.0 ÷ 164 = 0.189 = 18.9%
4 O’s = 4 x 16.0 = + 64.0 ÷ 164 = 0.390 = 39.0%
164
b) SnCl4
Sn = 118.7 ÷ 260.7 = 45.5%
4 Cl’s = 4 x 35.5 = + 142.0 ÷ 260.7 = 54.5%
260.7
Elements in the Universe:
% Composition by Mass
Earth’s Crust: % Composition by Mass
Entire Earth (Including Atmosphere):
% Composition by Mass
Human Body: % Composition by Mass
Determining the Empirical Formula for a Compound
•
The empirical formula for a compound is the simplest __________
whole
number __________
of the atoms in the compound.
ratio
Examples: H2O is the empirical formula for water.
C_______
1H2O1
is the empirical formula for glucose, C6H12O6.
Practice Problems: What is the empirical formula for the following
compounds? a) C6H6= CH
________
C4H7O
b) C8H14O2 = ________
c) C10H14O2 =C_________
5H7O
2
d) Ca5Br10 = CaBr
________
NO3
e) N3O9 = ________
Determining the Molecular Formula for a Compound
•
The molecular formula for a compound is either the same as the
empirical formula ratio or it is a “_________
whole # _________
multiple of this
ratio. It represents the true # of atoms in the molecule.
Examples: 1) H2O is the empirical & molecular formula for water.
2) CH2O is the empirical formula for sugar, ethanoic
acid, and methanol. The molecular formula for
6
glucose is C6H12O6, (___times
the empirical ratio!)
Practice Problems: (1) If the empirical formula for a compound is
CH2, which of the following is a possible molecular formula for the
compound? a) C8H16 b) C8H8 c) C4H2 d) C3H9
(2) If the empirical formula for a compound is C2H3, which of the
following is a possible molecular formula for the compound?
a) C2H6 b) C10H15 c) C6H12 d) C8H14
Determining the Molecular Formula for a Compound
•
Find the molecular formula for C2H7 if the molecular mass of the
compound is 93.0 g/mol.
C2H7 = 31.0 g/mol
C2H7 x 3
•
93.0 g/mol
31.0 g/mol
=3
= C6H21
Find the molecular formula for P2O5 if the molecular mass of the
compound is 283.88 g/mol.
P2O5 = 141.94 g/mol
P2O5 x 2 = P4O10
141.94 g/mol
=2
283.88 g/mol